Spectroscopic Notation

S,P,D,… -> orbital angular momentum l=0,1,2,…
Subscript on the right gives j
Superscript on the left gives multiplicity 2s+1

Example: $²$P $_{3 / 2}$ j = 3/2, s = 1/2, l = 1.

Hydrogen Atom

$²$P $_{1 / 2}$ state. l = 1, s = 1/2, j = 1/2. The total angular momentum is up along z-axis. $m_{j} = + \frac{1}{2}$ What is the probability to find electron with spin-down. So, $j_{1} = ℓ = 1$ , $j_{2} = s = \frac{1}{2}$ , $j = ℓ - s$ . The electron spin down relates to $m_{2} = m_{s} = - \frac{1}{2}$ .

$(| j m_{j} ⟩ =) | 1 / 2, 1 / 2 ⟩ = | 3 / 2 - 1, 3 / 2 - 1 ⟩ = \sqrt{\frac{1}{3 / 2}} | l, s - 1 ⟩ ⟩ - \sqrt{\frac{1 / 2}{3 / 2}} | l - 1, s ⟩ = \sqrt{\frac{2}{3}} | 1, - 1 / 2 ⟩ - \sqrt{\frac{1}{3}} | 0, 1 / 2 ⟩$ . So, the probability is 2/3.

We could relate this to spinors by using spherical harmonics with $ℓ$ and $m_{ℓ}$ . $(j m_{j} ⟩ =) | 1 / 2, 1 / 2 ⟩ = \sqrt{\frac{2}{3}} Y_{1}^{1} (Ω) \otimes | - ⟩ - \sqrt{\frac{1}{3}} Y_{1}^{0} (Ω) \otimes | + ⟩ ≐ \frac{1}{\sqrt{3}} [\begin{matrix} - Y_{1}^{0} \\ \sqrt{2} Y_{1}^{1} \end{matrix}]$ .

For $j_{1} = j_{2} = 1$ $| 21 ⟩ = | 2 (2 - 1) ⟩ = \frac{1}{\sqrt{2}} | 01 ⟩ + \frac{1}{\sqrt{2}} | 10 ⟩$

Chain of Atoms with Spins

More than 2 angular momenta. $\vec{J} = {\vec{J}}_{1} + {\vec{J}}_{2} + {\vec{J}}_{3} = ({\vec{J}}_{1} + {\vec{J}}_{2}) + {\vec{J}}_{3} = {\vec{J}}^{'} + {\vec{J}}_{3}$

Assume we have 3 spins. ${\vec{S}}_{1}, {\vec{S}}_{2}, {\vec{S}}_{3}$ . Assume 1,2 and 2,3 are neighbors.

$H = - 2 J ({\vec{S}}_{1} \cdot {\vec{S}}_{2} + {\vec{S}}_{2} \cdot {\vec{S}}_{3})$ . $s_{1} = s_{2} = s_{3} = 3 / 2$ . Let ${\vec{S}}_{T} = {\vec{S}}_{1} + {\vec{S}}_{2} + {\vec{S}}_{3}$ , ${\vec{S}}_{13} = {\vec{S}}_{1} + {\vec{S}}_{3}$ . Then, ${H, {\vec{S}}_{1}^{2}, {\vec{S}}_{2}^{2}, {\vec{S}}_{3}^{2}, {\vec{S}}_{13}^{2}, {\vec{S}}_{T}^{2}, {\vec{S}}_{T z}}$ So, $H = - 2 J \frac{{\vec{S}}_{T}^{2} - {\vec{S}}_{13}^{2} - {\vec{S}}_{2}^{2}}{2}$ . Thus, $H | s_{1} s_{2} s_{3} s_{13} s_{T} m_{T} ⟩ = - ℏ^{2} J (s_{T} (s_{T} + 1) - s_{13} (s_{13} + 1) - s_{2} (s_{2} + 1)) | s_{1} s_{2} s_{3} s_{13} s_{T} m_{T} ⟩ = J ℏ^{2} (s_{13} (s_{13} + 1) + 15 / 4 - s_{T} (s_{T} + 1)) | s_{1} s_{2} s_{3} s_{13} s_{T} m_{T} ⟩$

$s_{13} = s_{1} + s_{3}, | s_{1} - s_{3} | = 3, 2, 1, 0$

$s_T = $ see worksheet.